Please Help... Question 1

Discussion in 'Homework Help' started by saki, Jun 26, 2008.

  1. saki

    Thread Starter New Member

    Jun 26, 2008
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    Using the method of determinants, calculate the values of the circulating currents in the circuit shown in fig.8 below:
    Fig.8

    -j

    [​IMG][​IMG] 1+j3 V

    5V

    -j2

    +j3

    [​IMG]
     
  2. saki

    Thread Starter New Member

    Jun 26, 2008
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    0
    Check the attachment for the diagram
     
  3. thingmaker3

    Retired Moderator

    May 16, 2005
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    6
  4. saki

    Thread Starter New Member

    Jun 26, 2008
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    My attempt on the 8the question is this way:

    Let the current in loop having 5V be I1 and that in the other loop be I2.
    So we get,

    I1 + 3I1 + 3I2 + j3I1 –jI1 –jI2 -5 = 0
    (4+ j3 –j) I1 – jI2 – 5 = 0
    4+j2) I1 –jI2 = 5 …1

    Similarly, in Loop 2 we get,

    3I2 + 3I2 + 3I1 –j2I2 –JI2 – 1 – j3 = 0
    3I1 + (6 –j2 –j)I2 –(1+j3) = 0
    3I1 + (6-j3)I2 = 1+j3 ……2

    Taking both eq 1 and 2 into consideration we get,


    [ (4+j2) -j] [ I1 ] = 5
    3 (6-j3) I2 1+j3
     
  5. saki

    Thread Starter New Member

    Jun 26, 2008
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    Check the .doc attached here...

    I gave my attempt at question 8.

    For question 9 I know this is transformer issue and

    kVA of a single phase transformer = V x A
    and Z = V/I

    but I was told that I should consider the frequency and other factors into place so what do i do abt that particular point?

    The question does not talk anything about frequency or anything.. please help.
     
  6. JoeJester

    AAC Fanatic!

    Apr 26, 2005
    3,373
    1,159
    Your figure 9 mentions the frequency.
     
  7. saki

    Thread Starter New Member

    Jun 26, 2008
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    can you please guide me on how should i be solving it.

    I just need guidance on how to solve it.

    please help ASAP.
     
  8. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    saki,

    Here is the way I think prob 8 should be set up. You appear able to solve the two simultaneous linear equations. Ratch

    Code ( (Unknown Language)):
    1.   (4+j2)I1         -(3-j)I2       = 5
    2.   -(3-j)I1         (6-j3)i2       = -(1+j3)
     
  9. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    saki,

    The inductive reactance of a 0.1H coil at 50Hz is 2*3.14*50*0.1 = 31.42 ohms

    The mutual inductance of the transformer is M = 0.9*sqrt(L1*L2) = 0.9*0.1=0.09H, and the inductive reactance is 2*3.15*50*0.09 = 28.27

    Assume a load across A-B of ZL

    Setting up the two node equations

    Code ( (Unknown Language)):
    1.     (10+j31.42)I1     -(j28.27)I2    =  5
    2.    -(j28.27)I1        (10+j31.42+ZL) =  0
    Solving, we get

    I1 = [(10+j31.42+ZL)*5]/[(10+j31.42)(10+j31.42+ZL)+28.27^2]

    For open circuit ZL = infinite so I1 = 5/(10+j31.42) = .05-j0.14

    For short circuit ZL = 0 so I1 = 5/[(10+j31.42)+(28.27^2/(10+j31.42)] = 0.23-j0.11

    I will let you substitute 10-j10 for ZL and solve for I1

    Ratch
     
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